This is in calculus using optimization.
The distance between two points in the xy plane is:
D = [(x2-x1)^2 + (y2-y1)^2]^0.5
Let (x1,y1) be the point (3,0), then D is:
[(x2-3)^2 + (y2)^2]^0.5, or to save clutter:
D = [(x-3)^2 + y^2]^0.5
y = 8/x, so D = [(x-3)^2 + 64x^-2]^0.5
dD/dx = 0.5 [(x-3)^2 + 64x^-2]^-0.5 * [2(x-3) - 128x^-3]
D is at the minimum when dD/dx = 0, so solve for:
[2(x-3) - 128x^-3] = 0
x = 4, so y = 2.
The hyperbola is closest to (3, 0) at (4, 2) where the distance is the square root of 5.
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